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