Is g_{ab}u^au^b Constant for a Charged Particle in Curved Spacetime?

[latentcorpse]

Show that the equation of motion for a charged particle in curved spacetime implies
that $g_{ab}u^au^b$ is constant along a charged particle path with 4-velocity $u_a$. Hence this equation is consistent with the condition $g_{ab}u^au^b =-1$ arising in the denition of 4-velocity.

I don't know how to incorporate the charge aspect into the geodesic equation. Any ideas?

 

[stevenb]

I don't know how to incorporate the charge aspect into the geodesic equation. Any ideas?

If the problem is not specifying that the charge is in an electric or magnetic field, and is only under the influence of gravity (or a noninertial reference frame), then it seems to me the only relevance of the charge is that accellerating charge radiates electromagnetic waves. This implies a reaction force and kinetic energy loss for the particle, if the charge is accellerating in the chosen reference frame.

 

[arkajad]

Charge in geodesic equations can be incorporated in a 5-dimensional theory. Charge is then the conserved momentum related to winding up along the 5th dimension. In four dimensions, in Riemannian geometry, charged particles in an EM field do not propagate along geodesics.

 

[latentcorpse]

Charge in geodesic equations can be incorporated in a 5-dimensional theory. Charge is then the conserved momentum related to winding up along the 5th dimension. In four dimensions, in Riemannian geometry, charged particles in an EM field do not propagate along geodesics.

i don't understand. the question talks about 4-velocities so surely i amen't expected to work in 5 dimesnsions am I? I'm not even sure I know how to do that!

 

[arkajad]

I admit I misunderstood your question, I did not read it carefully. Can you write down the equation of charged particles that you are supposed to use and in the form that was given to you?

 

[latentcorpse]

I admit I misunderstood your question, I did not read it carefully. Can you write down the equation of charged particles that you are supposed to use and in the form that was given to you?

I wrote down everything that was in the question in my original post? DO you think some of the information is missing?

 

[arkajad]

You wrote: "Show that the equation of motion for a charged particle in curved spacetime ..."

So, the first thing to do is look somewhere in your notes or in the textbook in order to find this equation of motion. How else you will be able to show anything about it?

 

[latentcorpse]

You wrote: "Show that the equation of motion for a charged particle in curved spacetime ..."

So, the first thing to do is look somewhere in your notes or in the textbook in order to find this equation of motion. How else you will be able to show anything about it?

This definiteyl hasn't been covered in our notes. I have just checked through the index of Wald and all that comes up under "charged" is charged Kerr black holes and this part of the textbook seems a bit beyond me lol.

 

[arkajad]

Wald, "General relativity", Index "Lorentz force law"

 

[stevenb]

Wald, "General relativity", Index "Lorentz force law"

It's not clear to me why you guys are discussing Lorentz force law and electric or magnetic fields. The OP's first statement of the problem does not say that the charged particle is in any electric or magnetic field, but only that it is in curved spacetime. In such a case you would be interested in the radiation reaction force, and even that is only relevant in a reference frame that has the charge accelerating relative to it.

 

[arkajad]

It's not clear to me why you guys are discussing Lorentz force law and electric or magnetic fields.

Because the problem stated is a standard one, often posed and easy to answer when discussing Lorentz force law in GR. It is a good test problem.

 

[latentcorpse]

Wald, "General relativity", Index "Lorentz force law"

okay thanks. so we have

$u^a \nabla_a u^b = \frac{q}{m} F^b{}_cu^c$ as the Lorentz force equation.

i don't understand how $u^a \nabla_a u^b$ relates to $g_{ab}u^au^b$ though?

 

[arkajad]

Try to calculate

$$u^a\nabla_a$$ of $$g_{bc}u^bu^c$$. Use Leinbiz's dormula and the Lorentz force formula, and then the fact that $$F_{bc}$$ is antisymmetric.

 

[latentcorpse]

Try to calculate

$$u^a\nabla_a$$ of $$g_{bc}u^bu^c$$. Use Leinbiz's dormula and the Lorentz force formula, and then the fact that $$F_{bc}$$ is antisymmetric.

okay. can you tell me if I'm on the right track...

$u^ \nabla_a (g_{bc} u^b u^c)$
$=u^a ( \nabla_a g_{bc} ) u^b u^c + u^a g_{bc} ( \nabla_a u^b ) u^c + u^a g_{bc} u^b (\nabla_a u^c)$
the first term vanishes as $\nabla_a g_{bc}=0$ in order for parallel transport to hold
so we get
$=u^a \nabla_a u^b g_{bc}u^c + u^a \nabla_a u^c g_{bc} u^b$
=$\frac{q}{m} F^b{}_d u^d g_{bc} u^c + \frac{q}{m} F^c{}_d u^d g_{bc} u^b$

now i don't know how to use the antisymmetry to help me...

also i don't understand why i calculated $u^a \nabla_a ( g_{bc} u^b u^c)$ in the first place?

thanks!

 

[arkajad]

$=u^a \nabla_a u^b g_{bc}u^c + u^a \nabla_a u^c g_{bc} u^b$
=$\frac{q}{m} F^b{}_d u^d g_{bc} u^c + \frac{q}{m} F^c{}_d u^d g_{bc} u^b$

now i don't know how to use the antisymmetry to help me...

$${F^b}_d\, g_{bc}u^du^c=F_{cd}u^du^c=-F_{dc}u^du^c$$
$${F^c}_d\,g_{bc}u^du^b=F_{bd}u^du^b=F_{dc}u^du^c$$

also i don't understand why i calculated $u^a \nabla_a ( g_{bc} u^b u^c)$ in the first place?

thanks!

You want to check if $$u_a u^a$$ is constant along the trajectory. Since it is a scalar function, its derivative can be replaced by the covariant derivative. If s the length parameter then, for a scalar function f:

$$df(x(s))/ds=\partial_\mu f\, dx^\mu/ds=\partial_\mu f u^\mu=u^\mu\nabla_\mu f=u^a\nabla_a\,f$$

The rest you will probably be able to figure out yourself. (I was not paying attention to any convention of naming indices with Greek or Latin or Roman letters)

 

[latentcorpse]

$${F^b}_d\, g_{bc}u^du^c=F_{cd}u^du^c=-F_{dc}u^du^c$$
$${F^c}_d\,g_{bc}u^du^b=F_{bd}u^du^b=F_{dc}u^du^c$$
You want to check if $$u_a u^a$$ is constant along the trajectory. Since it is a scalar function, its derivative can be replaced by the covariant derivative. If s the length parameter then, for a scalar function f:

$$df(x(s))/ds=\partial_\mu f\, dx^\mu/ds=\partial_\mu f u^\mu=u^\mu\nabla_\mu f=u^a\nabla_a\,f$$

The rest you will probably be able to figure out yourself. (I was not paying attention to any convention of naming indices with Greek or Latin or Roman letters)

ok so we wanted to check $u_a u^a$ was constant along the path so surely we just had to verify that its covariant derivative vanishes i.e. that $\nabla_a ( g_{bc} u^b u^c)$? why do we have $u^a \nabla_a ( g_{bc} u^b u^c)$?

And do you have any suggestions for the second bit of my original question to do with the -1?

 

[arkajad]

Because $$u^a$$ is the tangent vector to the trajectory, so it is defined only on the trajectory. You can can calculate the derivative along the trajectory but not along arbitrary directions. The operator $$u^a\nabla_a$$ is sometimes called the "directional derivative" - along the path $$x^a(t)$$ for which $$u^a(t)=dx^\a(t)/dt$$, with t being any parameter that is being used to parametrize the path.

In fact I have said before that s was a length parameter. I have made a mistake here. We know, by assumption, that s is a parameter for which the Lorentz law holds. The whole purpose of this exercise was to show that s can be taken to be the length parameter!

 

[latentcorpse]

Because $$u^a$$ is the tangent vector to the trajectory, so it is defined only on the trajectory. You can can calculate the derivative along the trajectory but not along arbitrary directions. The operator $$u^a\nabla_a$$ is sometimes called the "directional derivative" - along the path $$x^a(t)$$ for which $$u^a(t)=dx^\a(t)/dt$$, with t being any parameter that is being used to parametrize the path.

In fact I have said before that s was a length parameter. I have made a mistake here. We know, by assumption, that s is a parameter for which the Lorentz law holds. The whole purpose of this exercise was to show that s can be taken to be the length parameter!

okay. i had seen that directional derivative stuff before but had forgotten it. thanks.

as for your 2nd paragraph here i don't really understand what you mean?

and do you have any ideas about the 2nd part of the original question with the -1 bit?

 

[arkajad]

So $$u^au^a$$ is constant along the trajectories. So, by multiplying the parameter by a constant it can be made into either 1 or -1, depending on its sign. Usually it is assumed that the physical trajectories are timelike and those spacelike are "unphysical".

 

[latentcorpse]

So $$u^au^a$$ is constant along the trajectories. So, by multiplying the parameter by a constant it can be made into either 1 or -1, depending on its sign. Usually it is assumed that the physical trajectories are timelike and those spacelike are "unphysical".

So there is no mathematical argument for why it should be -1 instead of +1? just the fact that it's unphysical to travel on a spacelike trajectory because it is outside the light cone?

 

[arkajad]

There is no mathematical argument that I know. After all there are spacelike geodesics that correspond to zero charge. Adding a small charge and a weak field to them should not lead to a mathematical catastrophe. But some physicists may suffer a heart attack.

 

[latentcorpse]

$${F^b}_d\, g_{bc}u^du^c=F_{cd}u^du^c=-F_{dc}u^du^c$$
$${F^c}_d\,g_{bc}u^du^b=F_{bd}u^du^b=F_{dc}u^du^c$$

just to check...

in the first line you used antisymmetry of F but what did you do to the second?

 

[latentcorpse]

never mind. i figured it out - i'd made a slight error.