Time-dependent Riemannian metric

[JohnSimpson]

Hi, I'm trying to attack a problem where the Riemannian metric depends explicitly on time, and is therefore a time-dependent assignment of an inner product to the tangent space of each point on the manifold.

Specifically, in coordinates I encounter a term which looks like

[tex]v^iv^j\frac{\partial g_{ij}(t,\gamma(t))}{\partial t}[/tex]

where gamma is a smooth curve on the manifold and v is an arbitrary element of the appropriate tangent space. I'd like to be able to write this object in a nice coordinate free way, but I can't quite see how to, since writing something like

[tex] \langle v_x,v_x \rangle_{\partial g/\partial t} [/tex]

doesn't make any sense, since the derivative of g does not necessairly satisfy the requirements of being an appropriate inner product...

Any insight on this, or on time dependent riemannian metrics in general, would help

 

[quasar987]

You can write this as

[tex]\dot{g}_{t,\gamma(t)}(v,v)[/tex]

for although dg/dt might not be an inner product, it is still a tensor.

 

[lavinia]

Hi, I'm trying to attack a problem where the Riemannian metric depends explicitly on time, and is therefore a time-dependent assignment of an inner product to the tangent space of each point on the manifold.

Specifically, in coordinates I encounter a term which looks like

[tex]v^iv^j\frac{\partial g_{ij}(t,\gamma(t))}{\partial t}[/tex]

where gamma is a smooth curve on the manifold and v is an arbitrary element of the appropriate tangent space. I'd like to be able to write this object in a nice coordinate free way, but I can't quite see how to, since writing something like

[tex] \langle v_x,v_x \rangle_{\partial g/\partial t} [/tex]

doesn't make any sense, since the derivative of g does not necessairly satisfy the requirements of being an appropriate inner product...

Any insight on this, or on time dependent riemannian metrics in general, would help

your expression looks like you are differentiating the length of a constant vector field along a curve. This is not an inner product.

 

[JohnSimpson]

your expression looks like you are differentiating the length of a constant vector field along a curve. This is not an inner product.

The length is nothing but the inner product of v with itself under g. Sorry, I tried to write something simplified when I probably should have just written out the whole thing. Let [tex] X : \mathbb{R} \times M \rightarrow TM [/tex] be a smooth time dependent vector field on a smooth Riemannian manifold (M,g), and let Y(t) be some other vector field along a path gamma, which is chosen to be an integral curve of X. I'm calculating

[tex]
\frac{d}{dt}\langle \langle Y,Y \rangle \rangle_g = \nabla_X \langle \langle Y,Y\rangle \rangle_{g(t,\gamma)} + ?
[/tex]

where ? is the other term arising from the fact that the metric itself varies with time independent from the fact that you're flowing along X. As was pointed out this "time partial derivative of g" will be an (0,2) tensor as well. I just need to make some sort of clear definition about what this object actually is independent of coordinates, and then show that in coordinates it becomes what i want it to be (namely, the quadratic form of the y's and the partial of g with respect to t). I suppose I could just define the coordinate independent object by its coordinate representation, or perhaps I could go back to the definitions of connections and try to formulate what i want from the ground up rigorously

 

[zhentil]

The space of symmetric, bilinear forms on a manifold is a vector space. Hence, it's its own tangent space. When you allow everything to vary with time, you're not calculating the derivative of the metric, you're calculating the derivative of a function. If you want the derivative, you'd have to "back out" all the extraneous derivatives.

As for the rest, I'm confused as to what you want. The derivative is a symmetric, bilinear form on the manifold. You don't need coordinates to define this. If you want to be ultra-precise, you form a difference quotient which lies for all non-zero h in a space. If the derivative exists, a fortiori it also lies in that space.

It's kind of like saying that pairing a one-form with a vector field is a function. We could go around in a circle, saying that we pick an integral curve of the vector field and calculate the function in coordinates. But then we're forgetting the fact that we defined a one-form precisely so that its pairing with a vector field is a function.

 

[JohnSimpson]

Thanks for the replies. Let me try to clear up where my confusion now lies.

For a path \gamma(t) on M, and a vector field \xi(t) along gamma, one can show using the properties of a Levi-Civita connection that

[tex]
\frac{d}{d t} \Big \langle \Big \langle \xi(t),\xi(t) \Big \rangle \Big \rangle_{g(\gamma(t))} = 2\Big \langle \Big \langle {\nabla}_{\gamma^\prime(t)} \xi(t),\xi(t) \Big \rangle \Big \rangle_{g(\gamma(t))}.
[/tex]

What I would like, for the case when the metric depends on time, is for a formula like the following to pop up instead

[tex]
\frac{d}{d t} \Big \langle \Big \langle \xi(t),\xi(t) \Big \rangle \Big \rangle_{g(t,\gamma(t))} = 2\Big \langle \Big \langle {\nabla}_{\gamma^\prime(t)} \xi(t),\xi(t) \Big \rangle \Big \rangle_{g(t,\gamma(t))} + \frac{\partial g}{\partial t}(t,\gamma(t))\cdot(\xi(t),\xi(t)).
[/tex]

where \partial g \partial t is, as you said, a symmetric bilinear form, that in components is nothing but a time and coordinate dependant matrix.

But I essentially just pulled that second formula out of a hat - as far as i can tell, it does not follow from the definitions I'm familiar with. So my belief is that I now need to go back to the definitions of a connection and of a metric connection, and modify them such that the second formula above logically follows. I hope that was clearer

 

[zhentil]

Aside from the confusing notation, why shouldn't that follow? Let f(x,y) be a function of two variables. What's d/dt f(t,t^2)? To put it more helpfully, perhaps, what if you're differentiating that expression with two fixed vectors at a specific point. I.e. let u and v be fixed vectors at a point p. Differentiate the function f(t)=g(t,u,v). That will give you the symmetric bilinear form piece.

Also, you should be more careful here. The equality you stated only holds if the connection is the levi-civita connection for the metric. It does not appear that the connection you're using is time-dependent.