# Gauge transforms, gravitational waves, and the speed of thought

The relation between Gauge theories and physical constants can be summed up as follows [1, verbatim]:

In gauge theories, the physical content is gauge invarient: the physical properties of a configuration do not change under a gauge transformation...

Gravitational waves are physical entities which can be measured both indirectly (e.g. the Hulse and Taylor binary pulsar [2]) and recently directly (i.e. LIGO).

Let the presence of (linearized) physical gravitational waves depends on the Gauge chosen - you need to use the harmonic gauge else you simply get waves that travel at the speed of thought. Eddington had similar arguments in is 1923 book [3].

I am confused, therefore about the application of gauges here. Our choice of gauge changes the physics. You can choose one gauge and get gravitational waves traveling at the speed of light, but choose another and you don't [3] i.e. our gauge conditions don't seem like gauge conditions, but cause actual physical difference between solutions. Please can someone explain the resolution and how it is applicable to think of this situation.

References

[1] Hemker, P.W. and Wesseling, P. eds., 2012. Multigrid Methods IV: Proceedings of the Fourth European Multigrid Conference, Amsterdam, July 6–9, 1993 (Vol. 116). Birkhäuser. (p63)

[2] Weisberg, J.M. and Taylor, J.H., 2005, July. The relativistic binary pulsar b1913+ 16: Thirty years of observations and analysis. In Binary Radio Pulsars (Vol. 328, p. 25).

[3] Eddington, A.S., 1930. Mathematical theory of relativity. General Books LLC.

• Do you know what a gauge is? – user121330 Mar 14 '17 at 15:32
• What is the the "speed of thought"? – user126422 Mar 14 '17 at 15:35
• 1. What is the "speed of thought"? 2. General relativity is not exactly a gauge theory like any other. You can reformulate it as a gauge theory of the Lorentz (or even the $\mathrm{GL}(N)$) group but the standard formulation is not a gauge theory. You can see this simply by noting that the gauge field should be the Christoffels yet outside the Palatini formulation the Christoffels are not considered dynamical. It is therefore unclear how the quote about gauge theories is supposed to be applicable to GR, in particular what the gauge transformations are. – ACuriousMind Mar 14 '17 at 15:35
• If I understand what you are asking, the same thing occurs in E&M. In the Coulomb gauge the potentials fill space instantaneously. Retardation becomes apparent only when calculating fields with the proper transverse current. – garyp Mar 14 '17 at 16:30
• @Quantumspaghettification I'm no expert, so I'll have to pass. But I'll make an additional observation: Gauges can be very non-intuitive. For example, the transverse current needed for the Coulomb gauge is not localized to the location of charge density. It actually fills all space. It may be good for calculations, but it's not so good for physical insight. – garyp Mar 14 '17 at 17:16

In this answer we are working in the linearized theory. Thus unless otherwise stated everything only holds to first order. Although I asked the question I have written the question in the style I would answer a question I had not asked (i.e. use of 'you' etc)

## The Meaning of a Gauge Transform in GR

Consider the two metrics: $$g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$$ $$g'_{\mu \nu}=\eta+h_{\mu \nu}-\partial_\mu \xi_\nu-\partial_\nu \xi_\mu$$ where $\xi_\nu$ is an arbitary but small covector. It turns out that $g_{\mu \nu}$ and $g'_{\mu \nu}$ in the same coordinates have the same Riemann tensor and therefore satisfy the same Einstein equations. This means they correspond to the same stress-energy tensor and therefore the same physical solution. Thus by making the transformation: $$h_{\mu \nu}\rightarrow h_{\mu \nu}-\partial_\mu \xi_\nu-\partial_\nu \xi_\mu$$ $$x_\mu \rightarrow x_\mu$$ we leave this physical situation unchanged. Such a transformation, in this context is called a gauge transformation.

## The Speed of Thought

You said:

Let the presence of (linearized) physical gravitational waves depends on the Gauge chosen - you need to use the harmonic gauge else you simply get waves that travel at the speed of thought.

This statement is not as general as you make out which can be proved by a simple argument:

Let $h_{\mu \nu}$ correspond to a physical gravitational wave (i.e. produces curvature) in the harmonic gauge then (to first order): $$h'_{\mu \nu}=h_{\mu \nu}-\partial_\mu \xi_\nu-\partial_\nu \xi_\mu$$ must also produce a physical gravitational wave but in general is not in the harmonic gauge.
Where your comment does apply is in the case of plane waves, as described here. If we take a solution of the form: $$h_{\mu \nu} =A_{\mu \nu} e^{ik_\rho x^\rho}$$ for some constant tensor $A_{\mu \nu}$. This only corresponds corresponds to a physical wave traveling at the speed of light if $h_{\mu \nu}$ is in the harmonic gauge, else it corresponds to a coordinate wave traveling at the 'speed of thought' (i.e. not the speed of light, and not a real wave). However, the metric with: $$h'_{\mu \nu}=A_{\mu \nu} e^{ik_\rho x^\rho}-\partial_\mu \xi_\nu-\partial_\nu \xi_\mu$$ corresponds to the same physical wave and is not in the harmonic gauge - we just can't write it as a plane wave.

## Concluding Remarks

The crux of your question lays in the statement:

You can choose one gauge and get gravitational waves traveling at the speed of light, but choose another and you don't [3] i.e. our gauge conditions don't seem like gauge conditions, but cause actual physical difference between solutions.

I hope I have convinced you that this statement is false. By changing gauge you change the form of the solution (i.e. from a nice plane wave to something nasty) but do not, and cannot change the physical content.