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We know that, according to EEP, an emitted photon would travel a curvilinear path in a uniformly accelerated compartment as well as a compartment that has been located in a uniform gravitational field (G-field). It is rational if we deduce that, for the observer in the G-field, the mentioned curvilinear path is independent of time, i.e., if the observer repeats the experiment at any arbitrarily chosen time, the emitted photon would travel the same path and hit the same point on the front wall of the compartment. Now, consider an inertial observer who wants to calculate photon's strike point on the front wall only for the uniformly accelerated compartment using special relativity. I want to know whether the strike point is also independent of time from the viewpoint of this observer.

Recall that the mentioned inertial observer is allowed to use special relativity for the accelerated compartment, knowing the facts that, e.g., the height of the compartment is Lorentz contracted considering its instant velocity and also its constant acceleration ${a_0}$ is reduced to ${a_0}(1-{v^2}/{c^2})^{3/2}$ from his own viewpoint, in order to calculate the height of the strike point for different time intervals. I have realized that calculations based on special relativity is not very easy for this case but I am eager to know whether the same calculations have been done as yet.

Moreover, in common literature on general relativity, when EEP is explained, a photon is emitted from outside the uniformly accelerated compartment passing through a tiny hole on the compartment's wall and then hits the front wall, whereas, for the equivalent case in G-field, the entire experiment is carried out inside the compartment. I want to know why this is the case because I think the equivalency for these cases can only work well for special conditions.

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  • $\begingroup$ Might be a silly comment, but are you sure you can use special relativity if the compartment in accelerated ? $\endgroup$ – Dimitri Jan 5 '17 at 9:16
  • $\begingroup$ Note that the EEP states that acceleration and gravity are locally equivalent i.e. they are indistiguishable within an infinitesimal volume element at the observer's position. $\endgroup$ – John Rennie Jan 5 '17 at 9:16
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    $\begingroup$ @Dimitri: yes, SR is the geometry of a flat spacetime and acceleration can be described perfectly well using SR. Students tend to be told otherwise when they start learning SR to stop their brains from melting, but it isn't true. $\endgroup$ – John Rennie Jan 5 '17 at 9:18
  • $\begingroup$ Yes, we can. Assume that you are an inertial observer in interstellar space away from any field. You see an accelerating compartment and you can easily apply special relativity. $\endgroup$ – Mohammad Javanshiry Jan 5 '17 at 9:19
  • $\begingroup$ Got it, thanks. I think it would be good to clarify this point in the question for other users. $\endgroup$ – Dimitri Jan 5 '17 at 9:20
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People learning Special Relativity are too often told that it is only a theory that works for inertial frames. This is not true. Special Relativity is simply a gravitational theory in flat space.

Let me clarify.

(The next few paragraphs are given for people just learning relativity and can be skipped for those familiar with the four-vector formalism.)

The essence of special relativity is the definition of the Lorentz-Invariant measure of proper time. For two events separated by a time $\Delta t$ and a spacial displacement $\Delta\textbf{x}$, we define

$$\Delta\tau^2=\Delta t^2-\Delta\textbf{x}^2$$

In units where $c=1$. To check that this is Lorentz-Invariant, a light pulse will also have $\Delta t=|\Delta\textbf{x}|$, and so $\Delta\tau=0$. Under Lorentz transformations, since the speed of light is kept constant, this is invariant. The invariance for other velocities can be easily checked.

The geometrical interpretation of this invariance is at the heart of SR and GR. If we define a four-vector as a vectoral object, $x^{\mu}$, with four components ($\mu=0,1,2,3$) such that $x^0=t$, $x^1=x$, $x^2=y$, and $x^3=z$. Then the infinitesimal form of the proper time is given by

$$\mathrm{d}\tau^2=\eta_{\mu\nu}\,\mathrm{d}x^{\mu}\,\mathrm{d}x^{\nu}$$

Where $\mu$ and $\nu$ are implicitly summed over and $\eta$ is a matrix with elements

$$\eta=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{pmatrix}$$

This is an example of what we call a metric tensor. For now it's not too important to understand the geometry of the metric tensor, just think of it as the matrix that calculates the "distance" between two objects.

Note, however, what happens when we change coordinates $x^{\mu}\to\xi^{\mu}$, where the $\xi$ coordinates depend arbitrarily on the $x$ coordinates. We have

$$\mathrm{d}\tau^2=\eta_{\mu\nu}\left(\frac{\partial x^{\mu}}{\partial\xi^{\rho}}\mathrm{d}\xi^{\rho}\right)\left(\frac{\partial x^{\nu}}{\partial\xi^{\sigma}}\mathrm{d}\xi^{\sigma}\right)\equiv g_{\rho\sigma}(\xi)\mathrm{d}\xi^{\rho}\mathrm{d}\xi^{\sigma}$$

Where we have defined a new beast, $g(\xi)$, which is now a metric tensor that depends on the position in spacetime. This is the fundamental object in GR (even though we're not doing GR!).

Let us actually compute an example of this. In spherical coordinates, we have $t=t$, $x=r\cos{\phi}\sin{\theta}$, $y=r\sin{\phi}\sin{\theta}$, and $z=r\cos{\theta}$. Transforming the coordinates, we have

$$\mathrm{d}\tau^2=\mathrm{d}t^2-\mathrm{d}r^2-r^2\left(\mathrm{d}\theta^2+\sin^2{\theta}\mathrm{d}\phi^2\right)$$

The coefficients of the differential coordinate changes depend on position!

There is another example (which is more relevant to what you want), called Rindler coordinates, in which the proper time element is given as

$$\mathrm{d}\tau^2=a^2x^2\mathrm{d}t^2-\mathrm{d}\textbf{x}^2$$

Although it isn't obvious, these coordinates describe constant acceleration $a$ in the $x$ direction, and are related to the intertial frame by the transformations

$$t\to\frac{1}{a}\text{arctan}\left(\frac{t}{x}\right),\hspace{0.5cm}x\to\sqrt{x^2-t^2},\hspace{0.5cm}y\to y,\hspace{0.5cm}z\to z$$

Since there is a coordinate transformation that relates this accelerating frame to a flat frame, it is perfectly compatible with special relativity.

Okay, this is a lot of talk about Special Relativity with no mention of how it relates to General Relativity. The basic idea is that in General Relativity there is not necessarily a transformation which takes the metric tensor $g$ to the flat metric $\eta$ at every point. It does, however, allow a transformation at any point $X$ such that $g(X)=\eta$ and $\partial_{\mu}g(X)=0$ (here, $\partial_{\mu}=\partial/\partial x^{\mu}$). This is what the equivalence principle really says.

Near a gravitational body, the metric tensor is approximately given by a Rindler metric (that is exactly the same as saying that near a gravitational body we can approximate the field as a constant acceleration). Since there is a transformation from Rindler to flat coordinates, we have that there is a set of coordinates (namely, free-falling coordinates) such that a gravitational field looks locally inertial!

I've taken a lot of time to explain to you why the equivalence principle works. That was a very long detour, and so now let's get to the heart of your question: the propagation of light in an accelerated frame vs a gravitational field.

I'll answer the last part of your question first. The results would have been identical if the laser pulse for the accelerated observer was created within the compartment or outside of it, as far as the intervals between emission and absorption are concerned.

As for the first part of your question: you consider a scenario in which a pulse is released periodically from a laser and you measure its path. In this coordinate system, the path itself is independent of time. However, in the inertial frame, since the actual laser itself is moving, the point at which the pulse lands is clearly time dependent. The compartment is not only moving, but it is contracted more and more the faster it goes. The interpretation that the inertial observer sees the compartment going faster and faster while it is getting shorter and shorter, so that the pulse is always hit at the same spot on the compartment.

I hope this is helpful. If anything is unclear (as it usually is in problems like this), feel free to ask questions!

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  • $\begingroup$ "A tourist in a powered interplanetary rocket feels "gravity." Can a physicist by local effects convince him that this "gravity" is bogus? Never, says Einstein's principle of the local equivalence of gravity and accelerations. But then the physicist will make no errors if he deludes himself into treating true gravity as a local illusion caused by acceleration. Under this delusion, he barges ahead and solves gravitational problems by using special relativity: if he is clever enough to divide every problem into a network of local questions, each solvable under such a delusion, ... $\endgroup$ – Robin Ekman Mar 16 '17 at 15:01
  • $\begingroup$ ... then he can work out all influences of any gravitational field. Only three basic principles are invoked: special-relativity physics, the equivalence principle, and the local nature of physics. To apply them however, he imposes a double task: (1) take spacetime apart into locally flat pieces (where the principles are valid), and (2) put these pieces back together in a comprehensible picture. To undertake this dissetion and reonstruction, to see curved dynamic spacetime inescapably take form, and to see the consequences for physics: that is general relativity." MTW, Box 6.1. $\endgroup$ – Robin Ekman Mar 16 '17 at 15:02
  • $\begingroup$ Thank you Bob for this detailed answer. Could you please specify if/where you use the clock postulate? (see my comment above). $\endgroup$ – user130529 Mar 17 '17 at 7:49
  • $\begingroup$ @Bob Knighton: Sorry, but I am not convinced. You just took a guess at the similarity of the hit points as the compartment accelerates. Precise calculations, using special relativity, are needed to validate this claim. Indeed, relevant calculations seem not to be very complicated and I'll soon do them on my own! $\endgroup$ – Mohammad Javanshiry Mar 17 '17 at 15:18
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    $\begingroup$ @Mohammad Javanshiry : precise calculation will indeed be very welcome. $\endgroup$ – user130529 Mar 17 '17 at 17:46

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