Even with Galilean relativity (with finite speed of information) the field will point at the real position. The arrow shooting argument 4:50 works in that case too. The arrow/light travels a straight line even in if the speed of arrows/light is not constant in every frame. However the angles will change in Galilean relativity: e.g. a light beam perpendicular to motion will be deflected backwards. And of course there is no Lorentz contractions effects.
Great video. I had been searching for an intuitive understanding of the Lienard-Wiechert field for a few months now. Before this video, I had resigned myself to trying to pull it out of the equations (which I might not have completed). Thank you for the insight :)
After watching this again, I realised that you stopped just short of showing how the Lienard-Wiechert field describes electromagnetic waves via acceleration. I'm particularly curious about how/if magnetism in EMR can be interpreted as a relativistic effect. Maybe that could be an option for a future video :)
Excellent question Dean. For some reason there are no discussions about this on the internet. I thought it over, and came to the conclusion that the magnetic part of EMR, can be understood as a correction to the red and blue shift of the radiation. I'm not done analyzing, but is seems to work out. I will consider it as a future topic.
hey this is a great video!!! it help me to understand a little better E fields of a point charge in motion, but i am curious about why you only covers EM topics in your videos? and can you tell me from which book did you get the heaviside equation for the E field of a point charge with velocity? and also which is the book you use for classical EM? thanks!
@@thesparetimephysicist9462 Since you like Maxwells equations so much, I make sure you have hear of Gravitoelectromagnetism :) en.wikipedia.org/wiki/Gravitoelectromagnetism
Sorry for a dumb question: I still dont understand how the 'duck shooting' applies to the electric field. In the case of the duck shooting, the arrow will travel along with the shooter, if he is in motion. But the emission of the field will travel independent of the speed of the charge(right?)... The field does not move according to the momentum of the emitting charge...? Or is the contraction of the field doing this precisely? Thanks
Hi Andre. The radiation moves at the speed of light no matter the velocity of the charge, so no extra momentum (except for Doppler shift of the radiation). If the charge is at rest the radial component is c, and the faster the charge moves, the slower the radial component becomes. The interesting thing about the example is the orientation of the arrow. It is not oriented in the direction of its motion when the charge is moving, but always towards the position of the charge. Does that answer the question?
@@thesparetimephysicist9462 Greetings, thanks for the reply. But, if the field radiates from the center of the charge, how could the orientation of the figurative "arrow" be aimed not at the center of the retarded position, (the point at which it was emitted), but at the real position? I guess what I fail to understand is how the 'arrow' is not be oriented in the direction of its motion if, as in the case of the field, its speed of travel is independent of the speed of the charge.
@@andremariaribeirodacunha3695 It is a fair question. In the case of electromagnetic radiation, it is in fact the case that we perceive the radiation as coming from the direction of its motion, and that is not always the direction of the source. en.wikipedia.org/wiki/Aberration_(astronomy) In the case of the "static" electric field surrounding the charge that I discuss in the video, we now experimentally that it will behave as i describe. The why question is more tricky. Bow hunting is an example of a system that behaves in the same way (except for the velocity addition), and can be used as an intuitive picture of it. You could also consider the electric potential. If we picture the potential as a landscape with the charge as a hilltop, the field is the gradient (slope) of the hill. For the "static" field you can imagine that the hill becomes length contracted, but that it otherwise follows the charge without deforming. If the charge changes its velocity, information about this change will propagate out with the speed of light. In this picture the answer to your question is that the gradient of the potential remains oriented towards the real position of the charge (at constant velocity). The retarded position is the position at which an update to the landscape was last sent.
I see the length contraction has the effect you describe. But it looks to me like there is another effect that is inline with the direction. en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Field_computation We have the value 1 - direction-to-target dot velocity. When velocity approaches lightspeed in the direction to target, this approaches zero. When velocity approaches lightspeed in the direction away from target, this approaches two. So appoaching the target, the force goes up with velocity, without limit. But fleeing from the target, the force goes down to limit 1/4 of the force at rest. Due to this term, not looking at the 1/alpha squared. Have I read this wrong?
Why is force large when two charges with squished fields pass each other moving to opposite directions? (Force is multiplied by gamma squared, probably) And why is force small when two charges with squished fields move side by side to same direction? (Force is divided by gamma, I'm sure of this)
Hi Jartas. So in both cases the angle is 90 degree, as I understand the question. If we consider the electric force of charge A on charge B, it will only depend on the velocity of charge A. Therefore the ELECTRIC force is the same in both cases, if the velocity of A is the same. (It will be amplified by a factor og gamma_A). What differs is the magnetic force. If the charges are moving in opposite direction the magnetic force will amplify the electric force, and vice versa. You could say that the magnetic accounts for the relative velocity between the two charges. Does that answer your question?
@@thesparetimephysicist9462 Oh yes magnetic fields. I forgot those exist, because I have believed this video saying that the magnetic effect is actually just the squishing effect: th-cam.com/video/1TKSfAkWWN0/w-d-xo.html I actually have my own idea why the electric force is small in a fast moving charged capacitor: The electric field has lost energy, which means that potential energy difference between plates is small, so force must be small. The energy loss we can see in moving and spinning flywheels too: The wheel has gained relativistic mass, gained rotational inertia, lost angular velocity, lost rotational energy, and retained angular momentum.
@@jartsajartsa6230 Yes, the magnetic force accounts for the "squishing" the affected charge experiences in its rest frame. Be aware that the explanation in the video you linked is not entirely correct. The magnetic force cannot be understood as an electrostatic effect. I explain it in this video. Most people are unaware of this. th-cam.com/video/rTE9gr-0Q0U/w-d-xo.html I see your line of thought, but don't think the electric field looses energy. Charge is conserved under relativistic transformation. It has to doo with the relativistic forcetransformations. I this case the force will transform F'=F/gamma. I think this has to doo with time being slowed down in the moving frame, which reduces the forces from the perspective of our reference frame. Whether it is related to the effect in the flywheel I don't know. Nice questions :-)
I wish I had more time to make them. The next one will be out in about 3 weeks. In it, I will explain a quite serious contradiction is electrodynamics that has not been pointet out before. I spent years with the problem, so i hope it will be worth the wait :-)
If you can plz explain the electric field of moving charge by using Heaviside equation,how field line contract in direction of motion and how the length of material bodies changes accordingly...I searched on TH-cam but couldn't find any video about this Heaviside ellipsoid
@@minhaskhan9164 Hi Minhas. In the equation i show, v is the velocity, and theta is the angle between the velocity vector and the field point at which you measure the field. You can try to plot the equation as function of those, to understand it better. I can't find any other videos. The change in the field is not the cause of the length contraction. It is space it self that contracts, and fields as well as particles follows. (The distance between two stars will change in the same way as the length of a rod would, if you accelerate relative to them. So it has nothing to doo with the matter or the forces in the matter) Hope that helps you.
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Even with Galilean relativity (with finite speed of information) the field will point at the real position. The arrow shooting argument 4:50 works in that case too. The arrow/light travels a straight line even in if the speed of arrows/light is not constant in every frame.
However the angles will change in Galilean relativity: e.g. a light beam perpendicular to motion will be deflected backwards. And of course there is no Lorentz contractions effects.
Great video. I had been searching for an intuitive understanding of the Lienard-Wiechert field for a few months now. Before this video, I had resigned myself to trying to pull it out of the equations (which I might not have completed). Thank you for the insight :)
You are so welcome Dean. I am happy that it is of use :-)
After watching this again, I realised that you stopped just short of showing how the Lienard-Wiechert field describes electromagnetic waves via acceleration. I'm particularly curious about how/if magnetism in EMR can be interpreted as a relativistic effect. Maybe that could be an option for a future video :)
Excellent question Dean. For some reason there are no discussions about this on the internet. I thought it over, and came to the conclusion that the magnetic part of EMR, can be understood as a correction to the red and blue shift of the radiation. I'm not done analyzing, but is seems to work out. I will consider it as a future topic.
hey this is a great video!!! it help me to understand a little better E fields of a point charge in motion, but i am curious about why you only covers EM topics in your videos? and can you tell me from which book did you get the heaviside equation for the E field of a point charge with velocity? and also which is the book you use for classical EM? thanks!
4:16 this is true for gravitational fields as well
I never considered that. Thanks
@@thesparetimephysicist9462 Since you like Maxwells equations so much, I make sure you have hear of Gravitoelectromagnetism :) en.wikipedia.org/wiki/Gravitoelectromagnetism
Sorry for a dumb question: I still dont understand how the 'duck shooting' applies to the electric field. In the case of the duck shooting, the arrow will travel along with the shooter, if he is in motion. But the emission of the field will travel independent of the speed of the charge(right?)... The field does not move according to the momentum of the emitting charge...? Or is the contraction of the field doing this precisely?
Thanks
Hi Andre. The radiation moves at the speed of light no matter the velocity of the charge, so no extra momentum (except for Doppler shift of the radiation). If the charge is at rest the radial component is c, and the faster the charge moves, the slower the radial component becomes. The interesting thing about the example is the orientation of the arrow. It is not oriented in the direction of its motion when the charge is moving, but always towards the position of the charge. Does that answer the question?
@@thesparetimephysicist9462 Greetings, thanks for the reply. But, if the field radiates from the center of the charge, how could the orientation of the figurative "arrow" be aimed not at the center of the retarded position, (the point at which it was emitted), but at the real position?
I guess what I fail to understand is how the 'arrow' is not be oriented in the direction of its motion if, as in the case of the field, its speed of travel is independent of the speed of the charge.
@@andremariaribeirodacunha3695 It is a fair question. In the case of electromagnetic radiation, it is in fact the case that we perceive the radiation as coming from the direction of its motion, and that is not always the direction of the source.
en.wikipedia.org/wiki/Aberration_(astronomy)
In the case of the "static" electric field surrounding the charge that I discuss in the video, we now experimentally that it will behave as i describe. The why question is more tricky. Bow hunting is an example of a system that behaves in the same way (except for the velocity addition), and can be used as an intuitive picture of it. You could also consider the electric potential. If we picture the potential as a landscape with the charge as a hilltop, the field is the gradient (slope) of the hill. For the "static" field you can imagine that the hill becomes length contracted, but that it otherwise follows the charge without deforming. If the charge changes its velocity, information about this change will propagate out with the speed of light.
In this picture the answer to your question is that the gradient of the potential remains oriented towards the real position of the charge (at constant velocity). The retarded position is the position at which an update to the landscape was last sent.
I see the length contraction has the effect you describe. But it looks to me like there is another effect that is inline with the direction.
en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Field_computation
We have the value 1 - direction-to-target dot velocity.
When velocity approaches lightspeed in the direction to target, this approaches zero.
When velocity approaches lightspeed in the direction away from target, this approaches two.
So appoaching the target, the force goes up with velocity, without limit.
But fleeing from the target, the force goes down to limit 1/4 of the force at rest.
Due to this term, not looking at the 1/alpha squared.
Have I read this wrong?
Why is force large when two charges with squished fields pass each other moving to opposite directions? (Force is multiplied by gamma squared, probably)
And why is force small when two charges with squished fields move side by side to same direction? (Force is divided by gamma, I'm sure of this)
Hi Jartas. So in both cases the angle is 90 degree, as I understand the question. If we consider the electric force of charge A on charge B, it will only depend on the velocity of charge A. Therefore the ELECTRIC force is the same in both cases, if the velocity of A is the same. (It will be amplified by a factor og gamma_A). What differs is the magnetic force. If the charges are moving in opposite direction the magnetic force will amplify the electric force, and vice versa. You could say that the magnetic accounts for the relative velocity between the two charges. Does that answer your question?
@@thesparetimephysicist9462
Oh yes magnetic fields. I forgot those exist, because I have believed this video saying that the magnetic effect is actually just the squishing effect:
th-cam.com/video/1TKSfAkWWN0/w-d-xo.html
I actually have my own idea why the electric force is small in a fast moving charged capacitor: The electric field has lost energy, which means that potential energy difference between plates is small, so force must be small.
The energy loss we can see in moving and spinning flywheels too: The wheel has gained relativistic mass, gained rotational inertia, lost angular velocity, lost rotational energy, and retained angular momentum.
@@jartsajartsa6230 Yes, the magnetic force accounts for the "squishing" the affected charge experiences in its rest frame. Be aware that the explanation in the video you linked is not entirely correct. The magnetic force cannot be understood as an electrostatic effect. I explain it in this video. Most people are unaware of this.
th-cam.com/video/rTE9gr-0Q0U/w-d-xo.html
I see your line of thought, but don't think the electric field looses energy. Charge is conserved under relativistic transformation. It has to doo with the relativistic forcetransformations. I this case the force will transform F'=F/gamma. I think this has to doo with time being slowed down in the moving frame, which reduces the forces from the perspective of our reference frame.
Whether it is related to the effect in the flywheel I don't know.
Nice questions :-)
Correction... If the the capacitor is at rest in the primed system it should be F=F'/gamma.
Waiting for the next video,your videos take a lot more time in uploading
I wish I had more time to make them. The next one will be out in about 3 weeks. In it, I will explain a quite serious contradiction is electrodynamics that has not been pointet out before. I spent years with the problem, so i hope it will be worth the wait :-)
If you can plz explain the electric field of moving charge by using Heaviside equation,how field line contract in direction of motion and how the length of material bodies changes accordingly...I searched on TH-cam but couldn't find any video about this Heaviside ellipsoid
@@minhaskhan9164 Hi Minhas. In the equation i show, v is the velocity, and theta is the angle between the velocity vector and the field point at which you measure the field. You can try to plot the equation as function of those, to understand it better. I can't find any other videos. The change in the field is not the cause of the length contraction. It is space it self that contracts, and fields as well as particles follows. (The distance between two stars will change in the same way as the length of a rod would, if you accelerate relative to them. So it has nothing to doo with the matter or the forces in the matter) Hope that helps you.
Yes thanks,I have the same idea but as a student and have poor knowledge I can't understand the Heaviside equations and what they are telling us.
Still waiting for the next video.....