Harmless Lorentz transformation question

[Physicsguru]

I have a harmless question which involves using the Lorentz transformations. Suppose that I am an observer in reference frame S, which is stipulated to be an inertial reference frame, and i am located at the origin of the frame (0,0,0).

A laser also located at (0,0,0) suddenly fires a photon along the x-axis of frame S, in the direction of increasing x coordinates.

Let t denote the time coordinate of inertial reference frame S.

At the moment the photon is emitted, let t=0.

Now, after some amount of time \Delta t has elapsed, the photon will have traveled some distance L, as measured by the X-axis ruler.

Now, let reference frame S` be attached to the photon, so that the photon is always at rest in reference frame S`. Furthermore, let the positive x` axis of frame S` coincide with the positive x-axis of frame S, let the positive y` axis of frame S` coincide with the positive y-axis of frame S, and let the positive z` axis of frame S` coincide with the positive z axis of frame S.

My question is simple.

If, after time amount of time \Delta t has elapsed according to some clock which is permanently at rest in S, the photon has coordinates (L,0,0) in frame S, what is my x` coordinate in frame S`?


For the sake of definiteness, suppose that exactly one second has ticked according to a clock at rest in frame S. Therefore, the location of the photon in frame S is given by (299792458 meters, 0,0).

Let (M,0,0)` denote the location of the origin of inertial reference frame S, in reference frame S`, at the instant that the clock at rest in frame S strikes one. Solve for M.

Thank You

Guru

 

[jtbell]

Now, let reference frame S` be attached to the photon, so that the photon is always at rest in reference frame S`.

There is no such reference frame, at least not one that can be related to other reference frames via the Lorentz transformation. The Lorentz transformation is not defined for v = c.

 

[Physicsguru]

There is no such reference frame, at least not one that can be related to other reference frames via the Lorentz transformation. The Lorentz transformation is not defined for v = c.

Obviously there are reference frames in which photons are at rest. Since you say that the Lorentz transformations won't allow me to transform coordinates correctly, can you tell me what coordinate transformations will?

In other words, just tell me what M is please.

Thank you

Guru

 

[JesseM]

Obviously there are reference frames in which photons are at rest. Since you say that the Lorentz transformations won't allow me to transform coordinates correctly, can you tell me what coordinate transformations will?

Thank you

Guru

You are free of course to use any coordinate transformation you like--the Galilei transform of Newtonian mechanics can give you a frame where the photon is at rest, for example. But all the most accurate known laws of physics are Lorentz-invariant, which means they will have the same form if you express them in terms of different Lorentzian coordinate systems, but they will not have the same form in different Galilean coordinate systems. This also insures that if you want your coordinate systems to have any physical meaning in terms of the readings on a network of rulers and clocks, you must use the Lorentz transform. This was actually how Einstein came up with the Lorentz transform, by imagining that each observer has a network of rulers and clocks at rest relative to himself throughout space, with each clock synchronized using the assumption that light travels at the same speed in all directions relative to himself. Then if each observer assigns space and time coordinates to different events by looking at the marking on the ruler right next to the event and the reading of the clock at that marker at the moment the event happened, the coordinates assigned to the same event by different observers will be related by the Lorentz transform. Check out my thread An illustration of relativity with rulers and clocks for some illustrations of how this works out.

 

[Hurkyl]

A coordinate chart in which a photon is at rest will not be an inertial reference frame.

I don't think it can even be a reference frame. As I recall, that requires four orthogonal, nondegenerate axes, and the time axis would be degenerate.

Furthermore, let the positive x` axis of frame S` coincide with the positive x-axis of frame S, let the positive y` axis of frame S` coincide with the positive y-axis of frame S, and let the positive z` axis of frame S` coincide with the positive z axis of frame S.

They can't coincide. I presume you merely meant that they're parallel?

In other words, just tell me what M is please.

Could be anything.

 

[quantumdude]

Obviously there are reference frames in which photons are at rest.

You seriously think that the Lorentz transformation can be used to boost to the rest frame of the photon, when the Lorentz transformation was derived from the assumption that no such frame exists?

Since you say that the Lorentz transformations won't allow me to transform coordinates correctly,

No one said that.

 

[Physicsguru]

A coordinate chart in which a photon is at rest will not be an inertial reference frame.

Can you prove the statement above?

Regards,

Guru

 

[Physicsguru]

I don't think it can even be a reference frame. As I recall, that requires four orthogonal, nondegenerate axes, and the time axis would be degenerate.

Would the time axis be degenerate if the Galilean transformations are correct in this instance?

Regards,

Guru

 

[Physicsguru]

They can't coincide. I presume you merely meant that they're parallel?

I meant coincide at the moment the laser fires the photon t=t`=0. Afterwards, the y, y` axis will no longer coincide of course, nor will the z, z` axes coincide, but the x,x` axes will still coincide.

And I don't want S` spinning relative to S, so stipulate that at all moments in time the y-axis is parallel to the y` axis, and the z axis is parallel to the z` axis.

Sorry if I didn't make that clear.

Regards,

Guru

 

[Physicsguru]

You seriously think that the Lorentz transformation can be used to boost to the rest frame of the photon, when the Lorentz transformation was derived from the assumption that no such frame exists?

Actually no I don't think so, I really just want to know what M is.

Regards,

Guru

 

[Physicsguru]

This was actually how Einstein came up with the Lorentz transform, by imagining that each observer has a network of rulers and clocks at rest relative to himself throughout space, with each clock synchronized using the assumption that light travels at the same speed in all directions relative to himself.

Where did he come up with that assumption... electrodynamics??

Is special relativity all a consequence of:

c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}

?

Regards,

Guru

 

[JesseM]

Can you prove the statement above?

Regards,

Guru

It depends how you define "inertial reference frame". Usually it's part of the definition that the laws of physics should work the same way in every inertial reference frame--if this is the case, then if all the most accurate laws are Lorentz-invariant, it shouldn't be hard to show that using some other transformation will always result in the laws of physics looking different in different frame's coordinate systems. On the other hand, if you just define "inertial reference frame" to mean that a point with a fixed set of space coordinates in one system (the origin, for example) should correspond to a point moving at constant velocity in other systems, then a coordinate system created by applying the Galilei transformation to some Lorentzian coordinate system could also be an "inertial frame" in this weaker sense.

 

[JesseM]

Where did he come up with that assumption... electrodynamics??

Is special relativity all a consequence of:

c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}

?

Regards,

Guru

Yes, basically. Since Maxwell's laws are not Galilei-invariant, then if they hold in one frame, they can't hold in a different frame related to the first by a Galilei transformation. So, physicists imagined that Maxwell's laws only hold exactly in one preferred frame, the rest frame of the "luminiferous ether", and that in other frames light would travel faster in one direction than the other. But Einstein showed that if each observer sees moving rulers shrink by \sqrt{1 - v^2/c^2}, and sees the length of time for a moving clock to tick extended by 1/\sqrt{1 - v^2/c^2}, then if each observer synchronizes his clocks by assuming light moves at the same speed in all directions relative to himself, then Maxwell's laws can hold exactly in every observer's reference frame.

 

[Physicsguru]

It depends how you define "inertial reference frame".

Well what is an inertial reference frame?

Regards,

Guru

 

[Physicsguru]

Yes, basically. Since Maxwell's laws are not Galilei-invariant, then if they hold in one frame, they can't hold in a different frame related to the first by a Galilei transformation.

I would appreciate if if you go a little slower, I'm not that smart.

Lets starte here: What is the quickest proof that Maxwell's laws aren't Galilei-invariant.

Regards,

Guru

 

[Hurkyl]

A coordinate chart in which a photon is at rest will not be an inertial reference frame.

Can you prove the statement above?

Yes. In inertial reference frames, light travels at c, thus cannot be at rest.

 

[krab]

Would the time axis be degenerate if the Galilean transformations are correct in this instance?

Your phrase "correct in this instance" betrays a lack of understanding of physical theory. Galilean tranformations are not correct, but a sufficiently good approximation when v<<c. Lorentz transformations are correct in all measured cases.

 

[JesseM]

I would appreciate if if you go a little slower, I'm not that smart.

Lets starte here: What is the quickest proof that Maxwell's laws aren't Galilei-invariant.

Regards,

Guru

Maxwell's laws say that all electromagnetic waves travel at the same velocity c in all directions. But using the galilei transformation, it's easy to see that if something has velocity v_1 along the x-axis in one frame, then in another frame which is moving at v_2 along the x-axis of the first frame, that same object must have velocity v_1 - v_2. So, if you use the Galilei transformation, an electromagnetic wave moving at velocity c along the x-axis in one frame could not be moving at c in other frames.

 

[dextercioby]

I would appreciate if if you go a little slower, I'm not that smart.

Lets starte here: What is the quickest proof that Maxwell's laws aren't Galilei-invariant.

Regards,

Guru

Perform a Galilei Transf.on the wavefunction and then try to see if the new wave function verifies the Galilei transformed D'Alembert equation.
Chose for simplicity
\psi=\psi(x,t)

Daniel.

 

[jcsd]

Well what is an inertial reference frame?

Regards,

Guru

A reference frame in which the two postulates of special relativity hold true. Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame.

 

[krab]

Lets starte here: What is the quickest proof that Maxwell's laws aren't Galilei-invariant.

Two electrons side-by-side. They repel each other, and so will start to move apart. Now observe them in a frame at constant velocity. They still electrically repel each other, but now in addition by Maxwell, because moving charges generate a magnetic field, the slightly attract each other as well. So the force of repulsion is reduced, and they move apart more slowly. In Galilean relativity, this is a contradiction.

 

[Physicsguru]

A coordinate chart in which a photon is at rest will not be an inertial reference frame.

Can you prove the statement above?

Yes. In inertial reference frames, light travels at c, thus cannot be at rest.

Hurkyl, using the definition of an inertial reference frame, can you derive the conclusion that the speed of light is c in all such frames? No. Instead, what you have to do, is show that if F is an inertial reference frame and E is one of Maxwell's equations then E is true in F.

In other words, you have to use Maxwellian electrodynamics to prove the following:

If F is an inertial reference frame and V denotes the speed of an "electromagnetic wave" or "photon" in frame F then V = \frac{1}{\epsilon_0 \mu_0}.

I would enjoy seeing such a proof, I anticipate that it will make some reference, either explicitly or implicitly to Faraday's law of induction. Also, I would like to caution you in advance, classical EM does not predict superconductivity, and don't reason on the converse.

Also, one more thing here is JCSD's quote:

A reference frame in which the two postulates of special relativity hold true. Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame.

Hurkyl... if JCSD is correct, then since any frame attached to a photon is moving at a constant speed in an inertial reference frame, it follows that the rest frame of a photon is an inertial reference frame.

My conclusion is this, there seems to be some confusion.
Kind regards,

Guru

 

[Physicsguru]

Maxwell's laws say that all electromagnetic waves travel at the same velocity c in all directions.

Jesse, see my challenge to Hurkyl. Also, if Maxwell's equations are unconditionally true, why don't they predict superconductivity?

Regards,

Guru

 

[jcsd]

As light has no frame of any kind, what I siad does not imply the rest frame of a photon is an inertial frame.

 

[Physicsguru]

Perform a Galilei Transf.on the wavefunction and then try to see if the new wave function verifies the Galilei transformed D'Alembert equation.
Chose for simplicity
\psi=\psi(x,t)

Daniel.

This is an interesting idea, but it does resort to using quantum mechanics, which has problems all its own. For example, just what exactly is the "wavefunction"? Since its complex valued, it cannot describe a physical wave.

Still I would like to see the mathematics behind this idea. Perhaps you could start me off with the idea, and I could finish?

D`Alembert's wave equation (1747):

U_t_t = c^2 U_x_x

Or equivalently


\frac{\partial^2 U}{\partial t^2} = c^2 \frac{\partial^2 U}{\partial x^2}

Where c denotes the wavespeed, and is a constant, and U(x,t) denotes the amplitude of the wave at position x, moment in time t.

I think my first step would be to rewrite the wave equation as follows:

\frac{\partial^2 U}{\partial t^2} - c^2 \frac{\partial^2 U}{\partial x^2} =0

Then I would factor it like so:

(\frac{\partial}{\partial t} + c \frac{\partial}{\partial x}) (\frac{\partial}{\partial t} - c \frac{\partial}{\partial x}) U = 0

At this point, I can clearly see that the equation is true if:


(\frac{\partial U}{\partial t} - c \frac{\partial U}{\partial x}) = 0

Or

(\frac{\partial U}{\partial t} + c \frac{\partial U}{\partial x}) = 0

Perhaps you can take it from here Dexter?

Regards,

Guru

 

[dextercioby]

Jesse, see my challenge to Hurkyl. Also, if Maxwell's equations are unconditionally true, why don't they predict superconductivity?

Regards,

Guru

How about another question:Why don't the very praised Maxwell equations predict the Bohm-Aharonov effect...??

Daniel.

P.S.The answers are identical to both questions:mine & yours.

 

[dextercioby]

This is an interesting idea, but it does resort to using quantum mechanics,

What QM are you dreaming about...??

For example, just what exactly is the "wavefunction"? Since its complex valued, it cannot describe a physical wave.

There's a well known fact that in classical electromagnetism,after solving the homogenous wave-equation,we pick the complex exponential solution for their easy usage.However,we are aware of the fact that both the physical fields and the ("imaginary" in the context of CED) potential are real functions and that's why we explicitely make the assumption of taking into account either the real part (cosine) or the imaginary part (sine,more frequent) of the complex exp...

D`Alembert's wave equation (1747):

U_t_t = c^2 U_x_x

Or equivalently


\frac{\partial^2 U}{\partial t^2} = c^2 \frac{\partial^2 U}{\partial x^2}

Where c denotes the wavespeed, and is a constant, and U(x,t) denotes the amplitude of the wave at position x, moment in time t.

I think my first step would be to rewrite the wave equation as follows:

\frac{\partial^2 U}{\partial t^2} - c^2 \frac{\partial^2 U}{\partial x^2} =0

Then I would factor it like so:

(\frac{\partial}{\partial t} + c \frac{\partial}{\partial x}) (\frac{\partial}{\partial t} - c \frac{\partial}{\partial x}) U = 0

At this point, I can clearly see that the equation is true if:


(\frac{\partial U}{\partial t} - c \frac{\partial U}{\partial x}) = 0

Or

(\frac{\partial U}{\partial t} + c \frac{\partial U}{\partial x}) = 0

Perhaps you can take it from here Dexter?

I have no interest of taking it from anywhere.I've done this calculation before knowing anything about Lorentz invariance of the wave-equation,which happened exactly 3 years ago...
Actually i proved it otherwise.I took the 3D wave equation and applied to every differential operator involving "t" or/and coordinate the Galilei transformation and showed that the new equation contained different terms than the one which would be exactly the initial wave-equation,only in the primed (transfomed) IRS.

Daniel.

P.S.I showed that using chain rule for partial derivatives only...

 

[Physicsguru]

As light has no frame of any kind, what I siad does not imply the rest frame of a photon is an inertial frame.

This is false. Since photons can change direction, they can be accelerated. That which accelerates has a center of inertial mass. Therefore, a photon has a center of mass. Since you can write formulas which are true in a reference frame in which the center of mass is at rest, you can write laws of physics which are true in reference frames in which a photon is at rest. So light, i.e. photons have frames.

Kind regards,

Guru

 

[dextercioby]

This is false. Since photons can change direction, they can be accelerated. That which accelerates has a center of inertial mass.

Explain that part,please... How do photons change direction and how are they accelerated...?

Daniel.

 

[Physicsguru]

Explain that part,please... How do photons change direction and how are they accelerated...?

Daniel.

Well for one, a mirror will reflect light that hits it.

Regards,

Guru

 

[dextercioby]

That is only at the surface,i won't ask u what trully happens to the photons...
What about the acceleration...?

Daniel.

 

[JesseM]

Jesse, see my challenge to Hurkyl. Also, if Maxwell's equations are unconditionally true, why don't they predict superconductivity?

Regards,

Guru

No one thinks Maxwell's laws are unconditionally true, but they were among the most accurate known laws in 1905. A hypothesis of relativity is that any new fundamental laws that come along in the future will also have the property of Lorentz-invariance, and this is indeed true; the laws of quantum electrodynamics (along with all other quantum field theories) are also Lorentz-invariant, for example.

 

[JesseM]

Hurkyl, using the definition of an inertial reference frame, can you derive the conclusion that the speed of light is c in all such frames?

What definition of an inertial frame? If an inertial frame is explicitly defined as one in which the laws of physics work the same way as in other inertial frames, then of course since light moves at c in slower-than-light inertial frames, any frame satisfying this defintion must also be a slower-than-light frame.

In other words, you have to use Maxwellian electrodynamics to prove the following:

If F is an inertial reference frame and V denotes the speed of an "electromagnetic wave" or "photon" in frame F then V = \frac{1}{\epsilon_0 \mu_0}.

I would enjoy seeing such a proof, I anticipate that it will make some reference, either explicitly or implicitly to Faraday's law of induction.

Are you just asking for a proof that Maxwell's laws prove than an electromagnetic wave (not a photon, they don't exist in classical electromagnetism) must always travel at \frac{1}{\epsilon_0 \mu_0}? If so, there's a proof at the bottom of this page, and another on this page.

Also, one more thing here is JCSD's quote:

A reference frame in which the two postulates of special relativity hold true. Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame.

Hurkyl... if JCSD is correct, then since any frame attached to a photon is moving at a constant speed in an inertial reference frame, it follows that the rest frame of a photon is an inertial reference frame.

He defined an inertial frame as one where the two postulates of relativity hold true (one of which is the postulate that all the fundamental laws of physics obey the same equations in all inertial frames), and this would not be true of any coordinate system you could define where the photon is at rest. He misspoke when he said any frame moving at constant velocity relative to another is also an inertial frame...this would be true in Newtonian physics, but in relativity it would only be true for coordinate systems that have a velocity v<c.

 

[Hurkyl]

Hurkyl, using the definition of an inertial reference frame, can you derive the conclusion that the speed of light is c in all such frames?

No. As you point out, I would invoke the principle of relativity. In particular, that Maxwell's equations are valid in any IRF.

I would enjoy seeing such a proof

It's rather straightforward, at least when using the differential form. You simply start with \partial^2 \vec{E} / \partial t^2 and make the appropriate substitutions using Maxwell's equations. (And the same for B).

Also, I would like to caution you in advance, classical EM does not predict superconductivity, and don't reason on the converse.

It doesn't predict photons either.

Hurkyl... if JCSD is correct, then since any frame attached to a photon is moving at a constant speed in an inertial reference frame, it follows that the rest frame of a photon is an inertial reference frame.

As I mentioned, a coordinate chart in which a photon is at rest can't be a reference frame because it has a degenerate time axis. (There is no unit vector in the time direction)

Since photons can change direction, they can be accelerated.

First off, photons don't even have 4-velocity vectors. Secondly, they can't be accelerated in SR.

Well for one, a mirror will reflect light that hits it.

The photon that comes back isn't the same photon that strikes the mirror.

 

[JesseM]

As I mentioned, a coordinate chart in which a photon is at rest can't be a reference frame because it has a degenerate time axis. (There is no unit vector in the time direction)

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules...but obviously the laws of physics would not work the same way in this coordinate system, and neither postulate of relativity would be satisfied, so if these conditions are part of the definition of "inertial frame", it wouldn't qualify as an inertial reference frame.

 

[jcsd]

This is false. Since photons can change direction, they can be accelerated. That which accelerates has a center of inertial mass. Therefore, a photon has a center of mass. Since you can write formulas which are true in a reference frame in which the center of mass is at rest, you can write laws of physics which are true in reference frames in which a photon is at rest. So light, i.e. photons have frames.

Kind regards,

Guru

Unfortunately that is wtrong for a lot of reasons:

1) photons don't accelrate but irrelvant

2) what is the inertial mass of a photon? You actually need a 3x3 matrix to represent somehting akin to inertial mass in special rleativty.

3) photons don't have rest mass and there is no frame in which they are at rest.

 

[jcsd]

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules...but obviously the laws of physics would not work the same way in this coordinate system, and neither postulate of relativity would be satisfied, so if these conditions are part of the definition of "inertial frame", it wouldn't qualify as an inertial reference frame.

The problem is a Galilean transformation is only the limit of a Lorentz transformation in special relativity, so it's not valid to use it to transform from one frame to the other. Hurkyl's reasiong is correct as it shows not only can you not define an inertial frame for light, but also that you can't even define a non-inertial frame for light.

 

[Hurkyl]

That'd be true if you tried to use the Lorentz transform to get such a coordinate system, but it would be possible to find a non-degenerate coordinate system in which every point is moving at constant velocity relative to Lorentzian frames and in which the photon is at rest, just by applying a Galilei transformation to a coordinate system constructed according to the relativistic rules

Not true: any vector lying along the time axis will have magnitude zero.

(In particular, note that the inner product is no longer given by a diagonal bilinear form)

 

[JesseM]

Not true: any vector lying along the time axis will have magnitude zero.

Well, it depends how you define the magnitude of vectors in the new coordinate system, if you define it in such a way as to insure it will agree with the magnitude of the same vector as seen in a valid inertial coordinate system, then of course this will be true. Is this all that the phrase "degenerate time axis" means? Any two events which have distinct coordinates in an inertial system will still have distinct coordinates in this non-inertial coordinate system, even if the two events lie along the worldline of a light ray.

(In particular, note that the inner product is no longer given by a diagonal bilinear form)

I don't know what the term "diagonal bilinear form" means...but like I said above, if you want to insure that the magnitude of a vector in the new coordinate system is the same as the magnitude of the same vector in an inertial coordinate system, you can't assume that the maginitude of two vectors (t0, x0, y0, z0) and (t1, x1, y1, z1) is given by -t0*t1 + x0*x1 + y0*y1 + z0*z1.

 

[JesseM]

The problem is a Galilean transformation is only the limit of a Lorentz transformation in special relativity, so it's not valid to use it to transform from one frame to the other.

It's "not valid" in the sense that the resulting coordinate system won't be an inertial one, but it will be an example of a non-inertial coordinate system which is moving at constant velocity relative to all inertial coordinate systems, which is why I think your earlier statement "Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame" was incorrect (unless by 'frame' you specifically meant 'inertial frame').

 

[jcsd]

Well, it depends how you define the magnitude of vectors in the new coordinate system, if you define it in such a way as to insure it will agree with the magnitude of the same vector as seen in a valid inertial coordinate system, then of course this will be true. But how do you define "degenerate"? After all, it is also true that certain vectors in inertial coordinate systems have magnitude zero, namely those that lie along the path of a light ray. I don't know what the term "diagonal bilinear form" means...but like I said above, if you want to insure that the magnitude of a vector in the new coordinate system is the same as the magnitude of the same vector in an inertial coordinate system, you can't assume that the maginitude of two vectors (t0, x0, y0, z0) and (t1, x1, y1, z1) is given by -t0*t1 + x0*x1 + y0*y1 + z0*z1. But likewise, you couldn't necessarily assume this in an accelerating coordinate system...would you say that accelerating coordinate systems are degenerate?

The magnitude of vectors is invariant, it's only decided by the metric tensor at that point. The metric tensor is a diagonal bilinear form.

Onm way of looking at it is that in every coordinate system we have four numbers to define each event, but the problem is that in our hypotehical refrenbce frma eof light one of those numbers is always zero, so we have three numbers to define each point in four dimensional space which cannot be done in a useful way.

The time basis vector in the rest frmae of an object is the unit vector tangent to it's worldline, in the case of light their is no unit vector tangent to it's wolrdline as any vector tangent to the worldline of light is null. No time basis vector, no coordinate system, no frame.

 

[jcsd]

It's "not valid" in the sense that the resulting coordinate system won't be an inertial one, but it will be an example of a non-inertial coordinate system which is moving at constant velocity relative to all inertial coordinate systems, which is why I think your earlier statement "Any frame traveling at constant velcoity rleative to an inertial frmae is also an inertial frame" was incorrect (unless by 'frame' you specifically meant 'inertial frame').

No it isn't, infact as a Galilean transformation presevres time we haven't done anything particularly interesting by performing a Galilean transformation, we're just using a different spatial coordinate system whose spatial origin (the points (t,0,0,0) vary with t). This is certainly not the refrence frame of light.

 

[JesseM]

No it isn't, infact as a Galilean transformation presevres time we haven't done anything particularly interesting by performing a Galilean transformation, we're just using a different spatial coordinate system whose spatial origin (the points (t,0,0,0) vary with t). This is certainly not the refrence frame of light.

I don't know what you mean by the phrase "the reference frame of light", I am certainly not claiming this coordinate system has any physical significance whatsoever, I'm just saying it's a non-inertial coordinate system which is moving at a constant velocity relative to all inertial frames, and in which an electromagnetic wave could be at rest.

 

[JesseM]

Onm way of looking at it is that in every coordinate system we have four numbers to define each event, but the problem is that in our hypotehical refrenbce frma eof light one of those numbers is always zero, so we have three numbers to define each point in four dimensional space which cannot be done in a useful way.

The time basis vector in the rest frmae of an object is the unit vector tangent to it's worldline, in the case of light their is no unit vector tangent to it's wolrdline as any vector tangent to the worldline of light is null. No time basis vector, no coordinate system, no frame.

Hmm, I hadn't thought about trying to express coordinates in terms of multiples of basis vectors...but why isn't it just as valid to express coordinates in terms of a coordinate transformation from some inertial system? Why can't you just pick an inertial system with coordinates x,y,z,t and then say:

x'=x - ct
y'=y
z'=z
t'=t

For any two points with distinct x,y,z,t coordinates, they'll be mapped to two points in this system with distinct x',y',z',t' coordinates.

 

[jcsd]

If anyhting it's a very odd inertial coordinate system. You can't go around labelling just anything speed d(x' + ct)/dt = c is still the velocity of an object with constant spatial coordinates in this coordinate system

 

[pervect]

One quick comment - there is such a thing in GR as a "null tetrad". This is something that I've been meaning to learn more about, it's closely related to Newmann-Penrose formalism and spinors. Instead of spacelike and timelike vectors, one has some (or perhaps even all) of the basis vectors being null vectors, which have a Lorentz interval of zero.

There's some discussion of this at

http://membres.lycos.fr/pvarni/dirac/node10.html

I've seen other examples where one mixes together null basis vectors with non-null vectors (i.e. ones that are space-like or time-like).

 

[jcsd]

Hmm, I hadn't thought about trying to express coordinates in terms of multiples of basis vectors...but why isn't it just as valid to express coordinates in terms of a coordinate transformation from some inertial system? Why can't you just pick an inertial system with coordinates x,y,z,t and then say:

x'=x - ct
y'=y
z'=z
t'=t

For any two points with distinct x,y,z,t coordinates, they'll be mapped to two points in this system with distinct x',y',z',t' coordinates.

Now I've had more time to think on it, what I guess you have there is a skew coordinate system (I.e. there will be off-diagonal entries in the metric in thta coordinate system). The reason I said it is inertial is thta it's time axis is the worldline of an inertial observer.

The thing with special relativity is that it is not general covraiant theory so arbitary coordinate transformations should be avoided as it means re-formulating the laws of physics. Though we can use other coordinate systems if we wish we should genrally base our our coordimante systems around Lorentz coordinates. Even non-inertial coordiante systems are generally going to be based around Lornetz coordinates by taking the worldline of our observer as as the time axis and then building the system using theat observers momentarily comoving inertial frames. In some ways the results can be unsatisfctory for non-inertial observers as non-inertial coordinate systems constructed in such a way are not generally well-behaved, BUT the observations of a non-inertial observer are not going to be 'well-behaved' so in this sense the coordinate system just refelcets the observations of the observer.

In special relativity the time basis vector is always the unit vector tangent to the observers worldline, so this is enough to exclude light from having a frame of reference.

I suppose i really didn't think about bases with null vectors, probably becasue of the unforunate terminology as the term null vector is often used to describe the additive identity (the zero vector) in a vector space and a set of n vectors containing the zero vector cannot span an n-dimensional space, wheras I guess null vectors can span Minkowski space.

 

[Physicsguru]

How about another question:Why don't the very praised Maxwell equations predict the Bohm-Aharonov effect...??

Daniel.

P.S.The answers are identical to both questions:mine & yours.

Short answer: Classical electrodynamics doesn't predict quantization of magnetic flux... a quantum effect exhibited by superconductors.

\Phi = magnetic flux = \oint \vec B \bullet d\vec a = n \Phi_0 = n(\frac{h}{2e})

Where \frac{h}{2e} is the magnetic flux quantum, n a positive integer, B the magnetic field.

Long answer:

Suppose that in some inertial reference frame, a charged particle is moving through a region of space at speed v in the frame, and that in the region of space local to the particle, there is a nonzero electric field OR a nonzero magnetic field, due to some non-local charge configuration nearby. According to classical EM, the particle will be accelerated in this inertial reference frame, and the total force acting upon this particle to accelerate it in this inertial frame, is the Lorentz force, which is given by:

\vec F = q[\vec E + \vec v \times \vec B]

Let us assume the classical relation above is a true statement, as regards the acceleration of the charged particle in the IRF, and the real force which gives it its acceleration in the IRF.

If the statement above is correct, then a charged particle which is moving through a region of space which is totally devoid of electric and magnetic fields (and other fields as well), will not experience an external force, and therefore, will not be accelerated. Hence, the charged particle will obey Galileo's Law of inertia (Newton's first law) in the inertial reference frame. That is, the particle will continue moving in a straight line at a constant speed.

Suppose now, that we design an experiment, where we have a charged particle pass near a solenoid. Using classical EM, the magnetic field exterior to the solenoid should be zero, and as long as the current in the solenoid is constant, there will also be no induced electric field either.

So then, we can control whether or not the solenoid is on or off in this experiment, but in either case, once the solenoid is on, or once the solenoid is off, there is no external B or E field due to the presence of the solenoid in our experimental setup, hence turning it on or off should not change the trajectory of charged particles which are passing near it.

Consider now a double slit experiment, in which electrons are impinging on the slits, and just behind the tiny wall separating the two slits, we have a tinier solenoid hidden there, which we can turn on, or off.

If the Lorentz force formula is correct, turning the solenoid on or off, should produce no effect on the trajectory of the electrons as they pass through the slits.

The experimental result is a diffraction pattern on the far wall, hence turning the solenoid on or off, should produce no noticable change in the observed diffraction pattern.

This is not what happens, instead the diffraction pattern is altered, the maxima and minima are switched, and the reason for this non-classical observation is called the Aharonov-Bohm effect.

Your question, is "why doesn't classical electrodynamics predict the effect?"

The answer is, because classical EM has the force which acts upon the charged particles as being caused solely by the electric or magnetic fields, when in an experiment in which the AB effect appears, since B=0 exterior to the solenoid whether the solenoid is on or off, it is the magnetic vector potential local to the particle which changes (and therefore is responsible for the force which gives rise to the AB effect), hence the Lorentz formula is wrong (not the whole story, because the potential gave rise to a force). In the next post, I will derive the formula for the magnetic vector potential.

Regards,

Guru

 

[dextercioby]

It's far easier the other way around,sorry to dissapoint you.

Maxwell equations are much more easily (and correctly when it comes to quantum behavior,like the B-A effect) expressible & integrable in terms of the potentials...

Daniel.

 

[dextercioby]

You didn't read my post.To me,what u did (namely extracting the \vec{A} once knowing the \vec{B}) means nothing.
Sorry.But that's the way i see it.

Daniel.

P.S.And yes,u still have some 6 equations to correct wrt the latex format.