General and Special Relativity Minkowski spaces

In summary: Regardless, in part b) you found that the christoffel symbol was zero. What does that imply about your answer to part c)? Does that give you something easy to solve for part d)?Yes, it means that the equation in part c) is solved for.
  • #1
kikitard
6
0

Homework Statement


In attached image

Homework Equations


?

The Attempt at a Solution


?

A start would be fantastic!
 

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  • #2
If you don't have ANY idea how to start ANY of these you might be in the wrong course. 3(a) just wants you describe the metric as a matrix. If (dt,dx,dy) is the vector ##dx^\mu## what matrix ##g_{\mu \nu}## makes ##dx^\mu dx^\nu g_{\mu \nu}## equal to the metric expression?
 
  • #3
I'm pretty sure I am in the wrong course, but it is required, which is why I've turned to the internet for help.

Am I correct in saying that the dxν = transpose of dxμ, and so gμν is the matrix with row vectors (-1,0,0) (0,1,0) (0,0,1), or am I off base?
 
  • #4
kikitard said:
I'm pretty sure I am in the wrong course, but it is required, which is why I've turned to the internet for help.

Am I correct in saying that the dxν = transpose of dxμ, and so gμν is the matrix with row vectors (-1,0,0) (0,1,0) (0,0,1), or am I off base?

That's a good start. Absolutely right. Now try 3(b).
 
  • #5
Using this formula (attached) can i retain the β in the formula, giving 1/2g[itex]\mu\beta((∂gβ\alpha/∂xβ)+(∂gββ/∂x\alpha)-(∂g\alpha\beta/∂xβ))[/itex]

we know gab=gba so the first and third terms in the brackets cancel
giving 1/2g[itex]\mu\beta(∂gββ/∂x\alpha)[/itex]
?
Would the partial derivative wrt xalpha be 0, as there are no xalpha terms contained in gββ?
 

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  • #6
kikitard said:
Using this formula (attached) can i retain the β in the formula, giving 1/2g[itex]\mu\beta((∂gβ\alpha/∂xβ)+(∂gββ/∂x\alpha)-(∂g\alpha\beta/∂xβ))[/itex]

we know gab=gba so the first and third terms in the brackets cancel
giving 1/2g[itex]\mu\beta(∂gββ/∂x\alpha)[/itex]
?
Would the partial derivative wrt xalpha be 0, as there are no xalpha terms contained in gββ?

That's hard to read. But all of the metric components are constant. So all of the partial derivatives of the metric are zero. So the Christoffel symbols are?
 
  • #7
Dick said:
That's hard to read. But all of the metric components are constant. So all of the partial derivatives of the metric are zero. So the Christoffel symbols are?

Ahh that makes sense, all partial derivatives are zero, so the Christoffel is also zero.

for c) I managed to get the attached image as the geodesic with affine λ
in d) i must solve this (presuming i did it correctly), could you point me towards the right method?
 

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  • #8
for c) You derived the geodesic equation in general it looks like. You got the right answer for that, but you can also try to find the specific case geodesic equation.

Regardless, in part b) you found that the christoffel symbol was zero. What does that imply about your answer to part c)? Does that give you something easy to solve for part d)?
 

1. What is the difference between general and special relativity?

General relativity is a theory of gravity that describes the relationship between matter and the curvature of space-time. Special relativity, on the other hand, is a theory that explains the behavior of objects moving at constant speeds in the absence of gravity.

2. How does Minkowski space relate to relativity?

Minkowski space is a mathematical concept that is used to describe the space-time geometry in special relativity. It combines the three dimensions of space with the dimension of time into a four-dimensional space-time continuum.

3. What is the significance of the speed of light in special relativity?

The speed of light, denoted as c, is the maximum speed at which all matter and information in the universe can travel. Special relativity states that the laws of physics should be the same for all observers moving at constant speeds, and the speed of light is a fundamental constant that is the same for all observers.

4. How does general relativity explain the bending of light?

In general relativity, light is affected by the curvature of space-time. As light travels through regions of high gravitational fields, it follows the curvature of space-time, appearing to bend from our perspective. This phenomenon is known as gravitational lensing.

5. Is time travel possible with relativity?

Time travel is a popular concept in science fiction, but it is not currently possible according to the laws of physics. While special relativity allows for time dilation, where time passes differently for objects moving at different speeds, it does not allow for traveling into the past. In general relativity, the possibility of time travel is still a topic of debate and has not been proven or disproven.

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