B Photography in 4D World: Time & Space

sonutabitha
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If time is a dimension, what would be the dimension of a photograph in such a space?
 
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Idealising the image as being one face of a physical piece of paper, it's a (2+1)d object, although there's nothing interesting about the time-like direction since (apart from any overall motion/deformation/damage/etc of the photograph) nothing changes.

Loosely, in Newtonian terms you regard a camera as mapping points (x,y,z) onto a plane (x,y) at a given time T-z/c (to allow for the finite speed of light) - it chooses a value of the time parameter and drops the z coordinate, in other words. In relativistic terms you'd regard it as taking points in the plane (x,y,z,T-z/c) and mapping them onto a plane (x,y) - so dropping the z and t coordinates. The only real difference is whether you regard time as a parameter or a dimension.
 
Ibix said:
Idealising the image as being one face of a physical piece of paper, it's a (2+1)d object, although there's nothing interesting about the time-like direction since (apart from any overall motion/deformation/damage/etc of the photograph) nothing changes.

Loosely, in Newtonian terms you regard a camera as mapping points (x,y,z) onto a plane (x,y) at a given time T-z/c (to allow for the finite speed of light) - it chooses a value of the time parameter and drops the z coordinate, in other words. In relativistic terms you'd regard it as taking points in the plane (x,y,z,T-z/c) and mapping them onto a plane (x,y) - so dropping the z and t coordinates. The only real difference is whether you regard time as a parameter or a dimension.
So are you saying that the photograph will be still 2 dimensional? I am considering time as a dimension not a parameter.
 
Objects are, in general, (3+1) dimensional. Idealising a photo as one surface of a piece of paper then it's (2+1) dimensional. There are two spacelike directions in the plane of the photo and it has extent in time. But the extent in time is boring because a photo is still only a recording of the light that struck it when it was exposed.

Why are you asking?
 
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sonutabitha said:
If time is a dimension, what would be the dimension of a photograph in such a space?

You are simply describing the world as it is now. Unless you add a temporal dimension into the photo, in which case you are describing a TV screen.
 
Algr said:
You are simply describing the world as it is now. Unless you add a temporal dimension into the photo, in which case you are describing a TV screen.
You need to be a bit careful about what you mean by "now" in relativity, which is why I answered as I did (at least at time of writing this thred is labeled A). In fact you are describing part of a null surface, the past light cone of the camera.

Otherwise, yes. Still curious in what context the OP wanted to know.
 
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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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