Zero areal velocity would mean that the body is stationary, so then what makes you think that it will be zero, and not the area of the cone? 

Dear Professor Vikash, 

Thank you on your question. The areal velocity in Euclidean 3-space  is the speed with which the surface area is covered by the position vector of the prticle in time t.  

The key question is : How can we find  an area of the lightcone, between   it's two  sufficiently close  lightlike rulings?  I know how to find  it in the case of Euclidean conical surface. An area between two sufficiently close rulings is approximated by the area of the triangle. Can we proceed in the same way in Minkowski 3-space? If we can , then I get the area is zero. 

@ Emilija, please explain what do you mean by "An area between two sufficiently close rulings is approximated by the area of the triangle. "

Dear Professors Cornwall and Pandey, 

Thanks on your opinion and reply. I know that hyperbolic geometry is not the same as Euclidean. In Euclidean case, an area of the conical surface between two sufficiently close rulings  r(t)=OA and  r(t+delta t)=OB ( when  delta  t tends to zero) and   where O is vertex of the cone,  is approximated by area of triangle OAB. We can not proceed in the same way in Minkowski space in the case of lightcone. Because  in Minkowski 3-space,  an area between two   sufficiently close rulings r(t)=OA and r(t+delta t)=OB is LIGHTLIKE, and it can not be approximated by area of triangle OAB, which is TIMELIKE. It shoulud be approximated by some lightlike area. The question is : Can we use an area of the lightlike triangle  which lies in the lightlike tangent plane of the lihghtcone? And how to obtain an area of the triangle lying in lightlike plane with degenerate metric? Such triangle may have  sides r(t) and v(t), where r(t) is the position vector of the particle,  v(t) is the speed of the base curve and t is time. 

Can you support your statement that rulings in Euclidean case is timelike and not lightlike?

Dear prof Vikash, 

I am sorry, but I do not understand your question. 

You have mentioned above ".... and it can not be approximated by area of triangle OAB, which is TIMELIKE. ". 

So how come the area of triangle OAB is TimeLike?

Triangle OAB lies in timelike plane, because the normal vector of that plane  alpha  is spacelike, as is the standard situation in Minkowski 3-space. The normal vector of the plane alpha  is equal to vector  

r(t) times r(t+ delta t). Then by using the indefinite  scalar product , say g, we get 

 g(r(t) times r(t+ delta t), r(t) times r(t+ delta t))= g( r(t), r(t+delta t))2 >0. 

This means that  the normal vector  r(t) times r(t+ delta t)  is spacelike. 

I forget to say that "times"  in the last answer means vector product of two vectors.  

Dear professor Robert, 

Thank you on your answer. The tangent plane of the null cone  is lightlike, so  the vectors A and B which span  such plane are spacelike and null (or vice versa) mutually orthogonal vectors.  Thus A x B =A, if A is null . This implies that

area of the ligthlike triangle over A and B   is equal to (1/2) || A x B||=0.

We also  may  conclude the following. 

If we approximate triangular element   ( say Delta A) in a null cone with :  

 Delta A =(1/2) r(t) x r(t+delta t), where r(t) and r(t+delta t) are two rullings,  then taking limes when delta t tends to zero, and dividing with delta t, we get 

dA/dt=A'(t)=(1/2) r(t) x r'(t)=(1/2) r(t), where ' denotes derivative with respect to time t. 

This means that areal velocity vector A'(t)=(1/2) r(t) is null vector. Its magnitude by definition  is areal velocity of the conical surface , in euclidean case. Can we conclude that areal velocity of the null cone  is zero, since it's areal velocity vector is null?